Struggling with Integrals? Don't Panic, Here's Some Tips!

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Homework Help Overview

The discussion revolves around the integral of sin²x with participants exploring various approaches to solve the problem. The subject area is calculus, specifically focusing on integration techniques.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss integration by parts and the use of trigonometric identities, including the double angle formulas for sine and cosine. There are attempts to simplify the integral using different methods, and some participants express frustration with the process.

Discussion Status

Some participants have offered tips and alternative methods for approaching the integral, suggesting that there may be simpler ways to solve it. There is an acknowledgment of the challenges faced, with some expressing a willingness to revisit their attempts.

Contextual Notes

There are indications of frustration and emotional responses to the problem, with participants sharing personal experiences related to studying integrals. The discussion reflects a mix of attempts and emotional reactions, highlighting the difficulties often encountered in calculus.

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Homework Statement



[itex]\int sin^{2}x dx[/itex]

Homework Equations





The Attempt at a Solution



I don't see why. Clearly it is designed for me to shoot myself before finishing.

u = (sinx)^2
du = 2sinxcosx dx, magically equal to sin2x dx
v = x
dv = dx

(-xcos2x/2) + (sin2x/4)

=

(-2xcos2x+sin2x)/(4) + C = wrong
 
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Since you just reviewed the double angle formula for sin, what's the double angle formula for cos?

cos(2x) = ? = ? = ?

Yes, there are three forms. One will work quite nicely here.
 
Firstly, let me apologize for my attitude. Secondly, thanks for the tip, I'll work it again and post back.
 
At times, getting frustrated can lead to good things.
 
[itex]\int sin^{2}x dx[/itex]

[itex]u = sin^{2}x[/itex]
[itex]du = sin2x dx[/itex]
[itex]v = x[/itex]
[itex]dv = dx[/itex]

[itex]xsin^{2}x - \int xsin2x dx[/itex]

[itex]u = x[/itex]
[itex]du = dx[/itex]
[itex]v = -\frac{cos2x}{2}[/itex]
[itex]dv = sin2x dx[/itex]

[itex]xsin^{2}x - [-\frac{xcos2x}{2} - \int -\frac{cos2x}{2} dx ][/itex]
[itex]xsin^{2}x - [-\frac{xcos2x}{2} + \frac{sin2x}{4}][/itex]

[itex]xsin^{2}x + \frac{xcos2x}{2} - \frac{sin2x}{4}[/itex]

[itex]\frac{2xsin^{2}x + xcos2x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{2xsin^{2}x + x(cos^{2}x - sin^{2}x)}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{2xsin^{2}x + xcos^{2}x - xsin^{2}x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{xsin^{2}x + xcos^{2}x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{x(sin^{2}x + cos^{2}x)}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{x}{2} - \frac{sin2x}{4}[/itex]
 
There's a much shorter way of doing this, following Sammy's hint.

sin2(x) = 1/2 * (1 - cos(2x))

This will give you an integral that you can do without having to resort to integration by parts.
 
Mark44's expression comes from solving the following for sin2(x) .

cos(2x) = 1 - 2 sin2(x) .

It's also true that

cos(2x) = cos2(x) - sin2(x)

cos(2x) = 2 cos2(x) - 1
 
1MileCrash said:
Firstly, let me apologize for my attitude.

th_smiley-bangheadonwall.gif


Welcome to the wonderful world of integrals! There are times when I want to kick small, defenseless animals because of this stuff. I usually listen to some thrash metal while studying math. I don't know why, but it helps!

Try to remember as many identities as you possibly can. It makes life a lot easier.
 

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